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Question

(A) Given that $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in an arithmetic progression,we have $\frac{2}{y} = \frac{1}{x} + \frac{1}{z}$.Given that $

Vedclass · Accepted Answer

$150$

Vedclass · Answer

(A) Given that $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in an arithmetic progression,we have $\frac{2}{y} = \frac{1}{x} + \frac{1}{z}$.Given that $x, \sqrt{2}y, z$ are in a geometric progression,we have $(\sqrt{2}y)^2 = xz$,which implies $2y^2 = xz$.Substituting $xz = 2y^2$ into the first equation: $\frac{2}{y} = \frac{x+z}{xz} = \frac{x+z}{2y^2}$,which simplifies to $x+z = 4y$.Given the equation $xy + yz + zx = \frac{3}{\sqrt{2}} xyz$,we can rewrite it as $y(x+z) + xz = \frac{3}{\sqrt{2}} xyz$.Substituting $x+z = 4y$ and $xz = 2y^2$: $y(4y) + 2y^2 = \frac{3}{\sqrt{2}} y(2y^2)$.$4y^2 + 2y^2 = \frac{3}{\sqrt{2}} (2y^3) \implies 6y^2 = 3\sqrt{2} y^3$.Since $y > 0$,we divide by $3y^2$: $2 = \sqrt{2}y$,so $y = \sqrt{2}$.Then $x+z = 4y = 4\sqrt{2}$,so $x+y+z = 5y = 5\sqrt{2}$.Finally,$3(x+y+z)^2 = 3(5\sqrt{2})^2 = 3(25 	imes 2) = 3(50) = 150$.

Let $0 < z < y < x$ be three real numbers such that $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in an arithmetic progression and $x, \sqrt{2}y, z$ are in a geometric progression. If $xy + yz + zx = \frac{3}{\sqrt{2}} xyz$,then $3(x + y + z)^2$ is equal to $............$.

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